guilherme: it couldn't be true for N because you could color each integer a distinct color.

But I agree that my hunch says it's true for R. I reformulate the question as follows: if you partition the reals into countably many subsets, can you always find four of them P1,...,P4 such that there are x1,...,x4 in the P's such that any two of them add up to the other two.

Hummm... I didn't realize the number of colors didn't have to be finite. I'm changing my vote to "a stupid question", which just means "a question I don't really care about".

The reason I don't care about it is that the problem is based on this extremely discontinuous use of the reals, where you can do things like assign each rational number a different color. You can't even color any interval, no matter how small, with the same color!

What the hell? Aren't you guys sick of coloring already given that everyone has been talking about coloring the graphs this way or that way, that now you have moved to coloring those innocent natural and real numbers? STOP DOING THAT!! All things colored not necessarily look good. At least leave N and R uncolored. Coloring, coloring, coloring, HUH!

If all color classes are measurable, then one of them must have positive measure, and so there has to be an interval within which that color class has density, say, 99%. Say that, after scaling, the interval is [0,1], then pick at random x in [0,1/4], w in [1/4,1/2] and epsilon in [0,1/2], and set z=x+epsilon and y = w+epsilon. Then by the union bound there is at least a 88% probability (or something like that) that all four points have the same color, and by construction x-z=w-y.

So if the answer is that a coloring exists, its definition must use the axiom of choice, because otherwise you can't construct non-measurable sets. (Under the proper assumptions, like ZF plus some cardinal assumption is consistent.)